If mass % of oxygen in monovalent metal carbonate is 48% ,then find the number of atoms of metal present in 5mg of this metal carbonate sample is (`N_(A) = 6.0 xx 10_(23)` )

Answer

To solve the problem step-by-step, we will follow the information provided in the question and the video transcript.

### Step 1: Define the Metal Carbonate
Let the monovalent metal be represented as \( M \). Since it is a carbonate, the formula for the metal carbonate can be written as \( M_2CO_3 \).

### Step 2: Calculate the Total Mass of Metal Carbonate
The molar mass of the metal carbonate \( M_2CO_3 \) can be expressed as:
\[ \text{Molar mass of } M_2CO_3 = 2M + 60 \]
where \( 60 \) is the molar mass of the carbonate ion \( CO_3^{2-} \).

### Step 3: Set Up the Mass Percentage Equation
According to the problem, the mass percentage of oxygen in the metal carbonate is given as \( 48\% \). The mass of oxygen in \( M_2CO_3 \) is \( 3 \times 16 = 48 \) g (since there are three oxygen atoms in the carbonate). Therefore, we can set up the equation:
\[ \frac{48}{2M + 60} \times 100 = 48 \]

### Step 4: Solve for M
From the equation, we can simplify:
\[ 48 = \frac{48}{2M + 60} \times 100 \]
This simplifies to:
\[ 2M + 60 = 100 \]
Now, solving for \( M \):
\[ 2M = 100 - 60 = 40 \\ M = 20 \text{ g/mol} \]

### Step 5: Calculate the Molar Mass of Metal Carbonate
Now substituting \( M \) back into the molar mass of \( M_2CO_3 \):
\[ \text{Molar mass of } M_2CO_3 = 2(20) + 60 = 40 + 60 = 100 \text{ g/mol} \]

### Step 6: Calculate Moles of Metal Carbonate in 5 mg
To find the number of moles in a 5 mg sample:
\[ \text{Moles of } M_2CO_3 = \frac{5 \times 10^{-3} \text{ g}}{100 \text{ g/mol}} = 5 \times 10^{-5} \text{ moles} \]

### Step 7: Calculate Moles of Metal
Since there are 2 moles of metal \( M \) for every mole of \( M_2CO_3 \):
\[ \text{Moles of metal } M = 2 \times 5 \times 10^{-5} = 10 \times 10^{-5} \text{ moles} \]

### Step 8: Calculate the Number of Atoms of Metal
Using Avogadro's number \( N_A = 6.0 \times 10^{23} \):
\[ \text{Number of atoms of metal} = N_A \times \text{moles of metal} = 6.0 \times 10^{23} \times 10 \times 10^{-5} = 6.0 \times 10^{19} \]

### Final Answer
The number of atoms of metal present in 5 mg of the metal carbonate sample is \( 6.0 \times 10^{19} \).

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